PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# Solution for F(X) = X/2+2/X, X>0 . - PUC Karnataka Science Class 12 - Mathematics

#### Question

f(x) = x/2+2/x, x>0 .

#### Solution

$\text { Given }: \hspace{0.167em} f\left( x \right) = \frac{x}{2} + \frac{2}{x}$

$\Rightarrow f'\left( x \right) = \frac{1}{2} - \frac{2}{x^2}$

$\text { For the local maxima or minima, we must have }$

$f'\left( x \right) = 0$

$\Rightarrow \frac{1}{2} - \frac{2}{x^2} = 0$

$\Rightarrow x^2 = 4$

$\Rightarrow x = 2\text { and } - 2$

$\text { Thus, x = 2 and x = - 2 are the possible points of local maxima or a local minima } .$

$\text { Since }x > 0, x = 2$

$\text { Now,}$

$f''\left( x \right) = \frac{4}{x^3}$

$\text { At }x = 2:$

$f''\left( 2 \right) = \frac{4}{\left( 2 \right)^3} = \frac{1}{2} > 0$

$\text { So, x = 2 is the point of local minimum } .$

$\text { The local minimum value is given by }$

$f\left( 2 \right) = \frac{x}{2} + \frac{2}{x} = 1 + 1 = 2$

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Solution F(X) = X/2+2/X, X>0 . Concept: Graph of Maxima and Minima.
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